How many units of electric power will be consumed by 4 motors of 0.5 HP each in 8 hours?
11.768 units
This question asks us to determine the total electric power consumed, measured in units (kilowatt-hours or kWh), by multiple electric motors operating over a specific duration. We need to calculate the combined energy usage of 4 motors, each rated at 0.5 HP, running for 8 hours.
First, it's essential to understand the relationship between Horsepower (HP), a unit of power, and Watts (W), the standard unit of power in the SI system. While different conventions exist, to match the expected result, we'll use the conversion factor for metric horsepower:
We can break down the calculation into the following steps:
We have 4 motors, and each motor has a power rating of 0.5 HP. The total power is the sum of the power of individual motors.
Total HP = Number of motors $ \times $ Power per motor
Total HP = $ 4 \times 0.5 \, \text{HP} $
Total HP = $ 2 \, \text{HP} $
Using the conversion factor 1 HP = 735.5 W:
Total Watts ($ W $) = Total HP $ \times $ 735.5 $ \, \frac{W}{HP} $
Total Watts = $ 2 \, \text{HP} \times 735.5 \, \frac{W}{HP} $
Total Watts = $ 1471 \, \text{W} $
Energy is calculated as Power multiplied by Time.
Energy (Wh) = Total Watts $ \times $ Time (in hours)
Energy (Wh) = $ 1471 \, \text{W} \times 8 \, \text{hours} $
Energy (Wh) = $ 11768 \, \text{Wh} $
One unit of electricity is equivalent to one kilowatt-hour (kWh). To convert Watt-hours to kilowatt-hours, we divide by 1000.
Energy (kWh) = Energy (Wh) $ \div 1000 $
Energy (kWh) = $ 11768 \, \text{Wh} \div 1000 $
Energy (kWh) = $ 11.768 \, \text{kWh} $
Therefore, the total electric power consumed by 4 motors of 0.5 HP each running for 8 hours is 11.768 units (kWh).
Which one of the following is the value of 1 KWh of energy converted into joules?
Which of the following is correct regarding electric power?
An elevator weighing 400 kg is to be lifted up at a constant velocity of 0.20 m/s. What would be the minimum horsepower of the motor to be used? (g = 9.8 m/s 2and there is no frictional loss)
An elevator weighing 500 kg is to be lifted up at a constant velocity of 0.25 m/s. What would be the minimum horsepower of the motor to be used? G = 9.8 m/s2 and there is no frictional loss.
Units of power is: